[Dan]

Assuming that the total volume does not significantly change, when mixing ROH with water:
Mixing aqueous KOH 1N and alcoholic KOH 1N,

In what ratio?

we get an initial mixture of KOH 1N per total volume unit that will immediately be transformed (ionic reaction) to (say) KOH 0.995 N and ROK 0.005 N per total volume and (say) KOH 1.98 N and ROK 0.02 N per ROH volume or a ratio 0.99/0.01

Please show your working. Writing things doesn't make them true.

The relative concentrations of KOH and ROK may change by further dilution with either water of ROH or their mixture but the concentration ratio 0.99/0.01 will remain constant, when calculated per ROH partition. 

Relative concentration and concentration ratio are the same thing.

As a reminder, the latter discussion refers to anhydrous ethanolic KOH.

So your example is irrelevant?

Can you show a valid calculation that supports your assertion that alkoxide formation in alcoholic KOH is negligible or not?

[pgk]

Sorry but I do not have any more time to spend for this discussion. Besides, the principal question of the discussion, has already been answered.